Compound (double) inequality Just want to understand why we have a specific approach to solving double inequalities for example separating the statement $-6 < 2x-4 < 12$ into two components. Working left two sides first and then the right two sides etc.
What is troubling me is this. The following statement is true.
$1 < 2 < 3$
In words, 3 is greater than 2 which is greater than 1. Let x = 2 therefore:
(x-1) < x < (x+1) 
The standard approach doesn't work anymore (I think) so if you work from left hand side,
x - 1 < x
x - x < 1
and whilst 0 is less than 1 which is true, there is no x variable to work with. Shouldn't this somehow return something like x > (0, 1, or any number less than 2) and x < (10, 20, or any number greater than 2)?
Why is that?
Or what am I doing wrong?
Thank You.
 A: Well, in your case, you assumed 
$(x-1) < x < (x+1)$ 
is true when $x=2$, but you also have to consider that $(x-1) < x < (x+1)$ is true for all real values of $x$, therefore it has no definite value which you can 'return' to.
You can try: 
$x=1: 0 < 1 < 2$
$x=100: 99 < 100 < 101$ and the list goes on
And, as a side note, we can see that the $x$'s cancel each other out, just like in an equation. For example, in
$x + y = x + 3$
$x$ can take on any value and the equation will remain solvable over $y$.
So: $(x-1) < x < (x+1) \rightarrow  -1 < 0 < 1$
A: I believe you are fundamentally misinterpreting how you are using variables in your example. In order to explain why, you must first understand what you are claiming. To say 

Shouldn't this somehow return something like x > (0, 1, or any number less than 2) and x < (10, 20, or any number greater than 2)?

is the same as saying
$x > 0, 1, 1.5, 1.9, 1.99, ...$
and
$x < 100, 10, 5, 2.5, 2.1, 2.01, ...$
We can equivalently say that $x \geq 2$ and $x \leq 2$. When you take these two inequalities together, you get $x = 2$.
However, $x-1 < x < x+1$ is a different claim altogether. Yes, $x = 2$ is one solution to this set of inequalities, but there are infinitely many other solutions. For example, if $x = 7$, then $6<7<8$. 
Therefore, you can play around with the inequalities, but you are not going to find the single solution $x = 2$ because the inequalities have more solutions, infinitely more in fact. 
